Homotopy between idempotents of small difference

Question

Let $A$ be a unital $C^*$-algebra. It is known that if $p$ and $q$ are projections in $A$ with

$$\|p-q\|<1,$$

then $p$ and $q$ are homotopic through a path of projections.

Question: Does a similar statement hold for idempotents? More precisely, if $e$ and $f$ are idempotents in $A$, does there exist $\delta>0$ such that $e$ and $f$ are homotopic through idempotents whenever $$\|e-f\|<\delta?$$

Answer 1

By Lemma (11.2.7) in (Rørdam, M.; Larsen, F.; Laustsen, N., An introduction to (K)-theory for (C^*)-algebras, London Mathematical Society Student Texts. 49. Cambridge: Cambridge University Press. xii, 242 p. (2000). ZBL0967.19001.), for every idempotent $e$ in $A$, one has that $$ \rho (e):= ee^*(1 + (e - e^*)(e^*-e))^{-1} $$ is a projection (self-adjoint idempotent) and $e\sim_h\rho (e)$ (meaning that $e$ and $\rho (e)$ are homotopic through a path of idempotents).

Thus, given $e$ and $f$ satisfying $\Vert e-f\Vert <\delta $ (the precise value of $\delta $ to be filled in later), we have that $e\sim_h\rho (e)$ and $f\sim_h\rho (f)$, so if we can manage to prove that $\rho (e)\sim_h\rho (f)$, we will get, by transitivity, that $e\sim_h f$. As noted in the OP, since $\rho (e)$ and $\rho (f)$ are projections, it would be enough to prove that $\Vert \rho (e)-\rho (f)\Vert <1$.

It is not difficult to see that the range of an idempotent element $e$ coincides with the range of $\rho (e)$, so $\rho (e)$ is in fact the orthogonal projection onto the range of $e$.

Given idempotents $e$ and $f$, let us henceforth write $E$ and $F$ for the ranges of $e$ and $f$, respectively, and by $p$ and $q$ the orthogonal projections onto $E$ and $F$, which amounts to saying that $p=\rho (e)$ and $q=\rho (f)$.

Define $$ \alpha (E, F) = \sup\{\text{dist}(x,F): x\in E,\ \Vert x\Vert \leq 1\}, $$ $$ \beta (E, F) = \sup\{\text{dist}(x,E): x\in F,\ \Vert x\Vert \leq 1\}. $$ and finally put $$ d(E, F) = \max\{\alpha (E, F),\beta (E, F)\}. $$

Lemma 1. We have $$ d(E,F)\leq \Vert e-f\Vert . $$ If moreover $e$ and $f$ are self-adjoint, then $$ \Vert e-f\Vert \leq 2d(E,F). $$

Proof. For $x$ in $E$ with $\Vert x\Vert \leq 1$, we have $$ \text{dist}(x,F) \leq \Vert x-f(x)\Vert = \Vert e(x)-f(x)\Vert \leq \Vert e-f\Vert , $$ so $\alpha (E, F)\leq \Vert e-f\Vert $, and it can be likewise proved that $\beta (E, F)\leq \Vert e-f\Vert $, whence $d(E, F)\leq \Vert e-f\Vert $.

Now assume that $e$ and $f$ are self-adjoint, so in particular $\Vert e\Vert \leq 1$ and $\Vert f\Vert \leq 1$. For every $x$ in $H$ with $\Vert x\Vert \leq 1$, we have that $e(x)\in E$ and $\Vert e(x)\Vert \leq 1$. Moreover, the element in $F$ closest to $e(x)$ is $f(e(x))$, so $$ \Vert e(x)-f(e(x))\Vert = \text{dist}(e(x), F) \leq \alpha (E,F)\leq d(E,F). $$ Taking the supremum for all $x$ in $H$ with $\Vert x\Vert \leq 1$, we deduce that $$ \Vert e-fe\Vert \leq d(E,F), $$ and a symmetric reasoning gives $\Vert f-ef\Vert \leq d(E,F)$, so also $$ \Vert f-fe\Vert = \Vert (f-ef)^*\Vert = \Vert f-ef\Vert \leq d(E,F). $$ This said we obtain $$ \Vert e-f\Vert = \Vert e-fe+fe-f\Vert \leq \Vert e-fe\Vert +\Vert fe-f\Vert \leq 2d(E,F). \tag*{$\blacksquare$} $$

Lemma 2. We have $$ \Vert p-q\Vert \leq 2\Vert e-f\Vert . $$

Proof. This follows from $$ \Vert p-q\Vert \leq 2d(E,F)\leq 2\Vert e-f\Vert . \tag*{$\blacksquare$} $$

Theorem. If $\Vert e-f\Vert <1/2$, then $e\sim_hf$.

Proof. By Lemma (2) we have $$ \Vert \rho(e)-\rho(f)\Vert =\Vert p-q\Vert \leq 2\Vert e-f\Vert <1, $$ so the conclusion follows as indicated above. $\qquad \blacksquare$

Answer 2

Supposing that $$ \Vert f-e\Vert < {1\over \Vert e \Vert +\Vert f \Vert}, \tag{*} $$ let $u=ef+(1-e)(1-f)$. Then $$ \Vert u-1\Vert = \Vert ef-e-f+ef\Vert= \Vert e(f-e) - (f-e)f \Vert \leq $$ $$ \leq \Vert e \Vert\Vert f-e\Vert + \Vert f-e \Vert\Vert f \Vert = (\Vert e \Vert +\Vert f \Vert)(\Vert f-e\Vert) <1. $$ This implies that $u$ is invertible. Moreover we have that $eu=uf$, whence $e=ufu^{-1}$.

Notice that the fact that $\Vert u-1\Vert <1$ implies not only that $u$ is invertible, but also that the power series defining the logarithm converges at $u$, so that $u=e^h$, for some $h$ in $A$. We then obtain a path of idempotents joining $f$ and $e$ by $$u_t=e^{th}fe^{-th}.$$

---

Remarks:

1. This does not provide a universal $\delta$, as required in the OP, but at least shows that being homotopically equivalent is invariant under small perturbations.

2. My impression is that there is no universal $\delta$ and I'll report back should I be able to prove it.

3. An elementary estimate shows that if $\Vert f-e\Vert <(4\Vert e\Vert )^{-1}$, then condition (*) above holds, so we deduce that $e$ is homotopically equivalent to every idempotent element in a neighborhood of $e$.